Permutation, Combination, Probability and Statistics

One question in the paper, but it can come from any of four topics: counting, probability, central tendency or spread. Learn the nPr/nCr formulas, the dice-and-cards facts, and the three averages.

The single question is usually the easiest in the paper if you know the vocabulary.

7.1 Fundamental principle, permutations and combinations

• Multiplication principle: m ways then n ways gives mn.
• {}^nP_r=\frac{n!}{(n-r)!}, {}^nC_r=\frac{n!}{r!(n-r)!}, {}^nC_r={}^nC_{n-r}, 0!=1, {}^nC_0=1.
• Letters of a word with repeats: \frac{n!}{p!\,q!\cdots} (BANANA: \frac{6!}{3!\,2!}=60).
• Two objects together: treat as one block, then multiply by 2!. Circular arrangements of n: (n-1)!.
• "At least one" is easiest as total minus none.

(a) Blaise Pascal, co-founder of probability

(b) Four dice: each face 1/6

(c) Two dice: 36 outcomes

7.2 Probability of events

• P(E)=\frac{\text{favourable}}{\text{total}}, 0\le P\le1, P(E)+P(E')=1.
• P(A\cup B)=P(A)+P(B)-P(A\cap B). Independent: P(A\cap B)=P(A)P(B).
• Facts: coin 2 outcomes; die 6; two dice 36 (sum 7 in 6 ways); pack of 52 cards = 26 red + 26 black, 13 per suit, 12 face cards, 4 aces; 3 coins have 8 outcomes.

7.3 Mean, median, mode and dispersion

• Mean =\frac{\sum x}{n}; median is the middle value of the ordered data (mean of the two middle values if n is even); mode is the most frequent value.
• Empirical relation: \text{mode}=3\,\text{median}-2\,\text{mean}.
• Range = max − min. Variance =\frac{\sum(x-\bar{x})^2}{n}, SD =\sqrt{\text{variance}}.
• Adding a constant to every value shifts the mean but leaves SD unchanged; multiplying by k multiplies the mean and SD by k.

7.4 Eight traps in counting, probability and statistics

These questions are short but unforgiving, because one wrong rule changes the answer completely. The eight traps below are the usual cause, and each is tested in the figures above.

1. "And" multiplies, "or" adds. If a girl can choose from 3 shirts and 4 skirts, there are 3\times4=12 outfits. If she picks either a shirt or a skirt, there are 3+4=7 choices. The fundamental principle works only when the choices are independent stages.
2. Order matters for permutations only. Choosing a president and a secretary from 5 is P(5,2)=20, but choosing a committee of 2 is C(5,2)=10 (Fig 7.1). Use the formulas P(n,r)=\frac{n!}{(n-r)!} and C(n,r)=\frac{P(n,r)}{r!}, and remember C(n,r)=C(n,n-r).
3. Repeated letters need division. The word BANANA has 6 letters with A three times and N twice, so the arrangements are \frac{6!}{3!\,2!}=60, not 720. For people around a round table the count is (n-1)!, because rotations are the same. Always check for repeats before writing n!.
4. Treat "together" as a block, and subtract for "never together". If two particular people must sit together in a row of 5, glue them: 4!\times2!=48. If they must not sit together, the count is 5!-48=72. For numbers made from digits, the first digit cannot be 0.
5. A probability lies between 0 and 1. The sample space of two dice has 6\times6=36 outcomes, not 12, and three coin tosses have 2^3=8 outcomes (Fig 7.2). Use P(A')=1-P(A). An answer such as \frac{7}{6} is always wrong.
6. Add the overlap back out. P(A\cup B)=P(A)+P(B)-P(A\cap B), as in the card example \frac{13}{52}+\frac4{52}-\frac1{52}=\frac4{13} (Fig 7.3). Mutually exclusive events have P(A\cap B)=0, so the last term vanishes. For "at least one", compute 1-P(\text{none}).
7. Independent is not the same as mutually exclusive. Independent events satisfy P(A\cap B)=P(A)P(B), while mutually exclusive events cannot happen together. Drawing two cards without replacement makes the draws dependent, so the second probability changes. Conditional probability is P(A|B)=\frac{P(A\cap B)}{P(B)}.
8. Sort first, and know what an outlier does. The median is the middle value of the sorted list, or the average of the two middle values when n is even (Fig 7.4). One extreme value moves the mean a lot but the median not at all. The variance is the mean of squared deviations, \sigma=\sqrt{\text{variance}}, adding a constant to every mark leaves the SD unchanged, and multiplying by k multiplies it by k.

7.4.1 Worked checks

Q. A committee of 3 is chosen from 5 men and 4 women. In how many ways does it contain at least one woman?

Answer. Total committees =C(9,3)=84. Committees with no woman =C(5,3)=10. So at least one woman: 84-10=74. Using the complement avoids three separate cases.

Q. Two dice are thrown. Find the probability that the sum is 7 or the dice show a double.

Answer. Sum 7 has 6 outcomes and doubles have 6, but no double sums to 7 (an odd total), so the overlap is 0. P=\frac{6+6}{36}=\frac13. Fig 7.2 shows both sets of cells.

Q. For the marks 4, 5, 5, 6, 7, 8, 8, 8, 9, 10 find the mean, median, mode and standard deviation.

Answer. Mean =\frac{70}{10}=7. The median is the average of the 5th and 6th values, \frac{7+8}{2}=7.5. The mode is 8. The squared deviations add to 34, so the variance is 3.4 and \sigma=\sqrt{3.4}=1.84.

Practice questions

Set A — Recall (20 questions)

1. The value of 5! isa24b60c120d720
2. The value of {}^5P_2 isa10b25c15d20
3. The value of {}^5C_2 isa10b20c5d25
4. The probability of getting a head on a fair coin isa1b\frac{1}{2}c\frac{1}{4}d0
5. The probability of a sure event isa0b\frac{1}{2}c2d1
6. The number of face cards in a pack of 52 isa4b16c12d13
7. The median of 3, 5, 7, 9, 11 isa5b7c9d35
8. The mean of 2, 4, 6, 8, 10 isa6b5c8d30
9. The mode of 2, 3, 3, 4, 5, 3, 6 isa2b4c5d3
10. {}^nC_0 equalsa0bnc1dn!
11. The number of arrangements of the letters of MAT isa6b3c9d12
12. P(A)+P(A') equalsa0b1c\frac{1}{2}dP(A)
13. The probability of getting a 6 when a die is thrown isa\frac{1}{3}b\frac{1}{2}c\frac{1}{6}d6
14. The number of outcomes when three coins are tossed isa3b6c9d8
15. The value of 0! isa1b0cundefineddinfinite
16. The standard deviation is the square root of theameanbvariancecrangedmode
17. The range of 4, 9, 15, 20 isa20b5c16d11
18. The value of {}^{10}C_8 isa90b80c10d45
19. The number of circular arrangements of n distinct objects isa(n-1)!bn!c\frac{n!}{2}d2n
20. The probability of drawing a red card from a standard pack isa\frac{1}{4}b\frac{1}{2}c\frac{1}{13}d\frac{3}{4}

Set B — Understanding (12 questions)

1. The number of distinct arrangements of the letters of BANANA isa720b120c30d60
2. A committee of 3 is chosen from 8 people. The number of ways isa56b336c24d28
3. Two dice are thrown. The probability that the sum is 7 isa\frac{1}{12}b\frac{1}{36}c\frac{1}{6}d\frac{1}{3}
4. For independent events with P(A)=0.4 and P(B)=0.5, P(A\cap B) isa0.9b0.2c0.1d0.45
5. If P(A)=0.5, P(B)=0.4 and P(A\cap B)=0.2, then P(A\cup B) isa0.7b0.9c0.1d0.2
6. The mean of five numbers is 12. A sixth number 18 is added. The new mean isa12b12.5c14d13
7. If the median is 20 and the mean is 18, the mode by the empirical relation isa22b26c24d21
8. Every value in a data set is increased by 5. The standard deviationaincreases by 5bis unchangedcdoublesdbecomes zero
9. The number of words that can be formed by rearranging all letters of MATH isa12b16c6d24
10. The probability of two heads in two tosses of a fair coin isa\frac{1}{4}b\frac{1}{2}c\frac{1}{3}d\frac{3}{4}
11. Five people stand in a row with two particular people always together. The number of arrangements isa24b48c72d120
12. The standard deviation of 2, 4, 4, 4, 5, 5, 7, 9 isa1b4c2d5

Set C — Application (8 questions)

1. The number of 3-digit numbers that can be formed from the digits 1 to 5 without repetition isa60b125c10d15
2. Two cards are drawn from a pack of 52. The probability that both are aces isa\frac{1}{169}b\frac{1}{26}c\frac{1}{221}d\frac{1}{13}
3. A committee of 3 is chosen from 4 men and 3 women with at least one woman. The number of ways isa35b30c34d31
4. The probability of at least one head in three tosses of a fair coin isa\frac{3}{8}b\frac{7}{8}c\frac{1}{8}d\frac{1}{2}
5. The variance of 40, 50, 60, 70, 80 isa200b100c50d14
6. If {}^nC_2=28, then n isa6b7c9d8
7. If {}^nP_2=42, then n isa6b7c8d9
8. A number is chosen at random from 1 to 20. The probability that it is a multiple of 3 or 5 isa\frac{1}{2}b\frac{2}{5}c\frac{9}{20}d\frac{11}{20}